All categories

3 years ago

Could someone help me eliminate some of these answer choices?

2 answers:

8 0

In order to be processed by the brain, sound waves are converted into :
A. Electrical signal waves
In order to be carried out by our neurons, information has to be transferred as electrical wave through our brain nerves

hope this helps

Nuetrik [128]3 years ago

8 0

Explanation :

There are many types of waves like transverse waves or longitudinal waves. The sound is a longitudinal wave and light is a transverse wave. The longitudinal waves propagate in the form of compression and rarefaction.

Compression is the position where the particles are present in bulk i.e have maximum pressure or density. On the other hand, rarefaction is the position where the particles are far apart from each other.

Due to this pressure difference, noise is created into our eardrum that moves to the brain through the auditory nerve. This noise is in the form of electrical signal waves.

So, the sound waves are converted finally into electrical signal waves in order to be processed by the brain.

Hence, the correct option is (A).

You might be interested in

A force in the +x -direction with magnitude F(x)=18.0N−(0.530N/m)x is applied to a 7.90 kg box that is sitting on the horizontal

dsp73

Answer:

$v\approx 8.570\,\frac{m}{s}$

Explanation:

The equation of equlibrium for the box is:

$\Sigma F_{x} = 18\,N-(0.530\,\frac{N}{m} )\cdot x = (7.90\,kg)\cdot a$

The formula for the acceleration, given in $\frac{m}{s^{2}}$ , is:

$a = \frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg}$

Velocity can be derived from the following definition of acceleration:

$a = v\cdot \frac{dv}{dx}$

$v\, dv = a\, dx$

$\frac{1}{2}\cdot v^{2} = \int\limits^{17\,m}_{0\,m} {\frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg} } \, dx$

$\frac{1}{2}\cdot v^{2} =\frac{18\,N}{7.90\,kg} \int\limits^{17\,m}_{0\,m}\, dx - \frac{0.530\,\frac{N}{m} }{7.90\,kg} \int\limits^{17\,m}_{0\,m} {x} \, dx$

$\frac{1}{2}\cdot v^{2} = (2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}$

$v =\sqrt{2\cdot[(2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}] }$

The speed after the box has travelled 17 meters is:

$v\approx 8.570\,\frac{m}{s}$

3 0

2 years ago

A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.15

Oksanka [162]

1) 5.5 N

When the ball is at the bottom of the circle, the equation of the forces is the following:

$T-mg = m\frac{v^2}{R}$

where

T is the tension in the string, which points upward

mg is the weight of the string, which points downward, with

m = 0.158 kg being the mass of the ball

g = 9.8 m/s^2 being the acceleration due to gravity

$m \frac{v^2}{R}$ is the centripetal force, which points upward, with

v = 5.22 m/s being the speed of the ball

R = 1.1 m being the radius of the circular trajectory

Substituting numbers and re-arranging the formula, we find T:

$T=mg+m\frac{v^2}{R}=(0.158 kg)(9.8 m/s^2)+(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=5.5 N$

2) 3.9 N

When the ball is at the side of the circle, the only force acting along the centripetal direction is the tension in the string, therefore the equation of the forces becomes:

$T=m\frac{v^2}{R}$

And by substituting the numerical values, we find

$T=(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=3.9 N$

3) 2.3 N

When the ball is at the top of the circle, both the tension and the weight of the ball point downward, in the same direction of the centripetal force. Therefore, the equation of the force is

$T+mg=m\frac{v^2}{R}$

And substituting the numerical values and re-arranging it, we find

$T=m\frac{v^2}{R}-mg=(0.158 kg)\frac{5.22 m/s)^2}{1.1 m}-(0.158 kg)(9.8 m/s^2)=2.3 N$

4) 3.3 m/s

The minimum velocity for the ball to keep the circular motion occurs when the centripetal force is equal to the weight of the ball, and the tension in the string is zero; therefore:

$T=0\\mg = m\frac{v^2}{R}$

and re-arranging the equation, we find

$v=\sqrt{gR}=\sqrt{(9.8 m/s^2)(1.1 m)}=3.3 m/s$

7 0

2 years ago

Which of the following is NOT common of elite Shang burials?

Zigmanuir [339]

Answer:

Large above ground mausoleums were not common in the elite Shang burials.

Explanation:

Large, above the ground mausoleums were not common so the answer is option B.

6 0

2 years ago

1)the car's engine power is 44000W. Explain this number in a physical sense

Ratling [72]

Answer:

1) It expresses the rate (top speed) at which it can move with time.

2) P = 20 W

3) h = 18 km

Explanation:

1) Power is the rate of transfer of energy.

⇒ Power = $\frac{Energy(or workdone)}{Time}$

i.e P = $\frac{E}{t}$

Thus a car's engine power is 44000W implies that the engine of the car can propel the car at this rate. This expresses the rate (top speed) at which it can move with time.

2) m = 400g = 0.4 kg

t = 20 s

h = 100m

g = 10 m/ $s^{2}$

P = $\frac{mgh}{t}$

= $\frac{0.4*10*100}{20}$

= $\frac{400}{20}$

P = 20 W

3) u = 600 m/s

g = 10 m/ $s^{2}$

From the third equation of free fall,

$V^{2}$ = $U^{2}$ - 2gh

V is the final velocity, U is the initial velocity, h is the height.

0 = $(600)^{2}$ - 2 x 10 x h

0 = 360000 - 20h

20h = 360000

h = $\frac{360000}{20}$

= 18000

h = 18 km

The maximum height of the bullet would be 18 km.

3 0

2 years ago

What change occurs to the mass of an object when an unbalanced force is applied to it

strojnjashka [21]

The mass of an object always stays the same since it is really just the amount of matter in an object so no matter the force applied, as long as the object does not lose or gain matter, the object stays the same

5 0

1 year ago